Solving $\cos A + \cos B + \cos C = 0$ and $\sin A + \sin B + \sin C = 0$ using complex numbers My teacher gave this question to me while teaching complex numbers:

If $\;\sin A\!+\!\sin B\!+\!\sin C=\cos A \!+\!\cos B\!+\!\cos C=0$, then find:

*

*$\cos 2A + \cos 2B + \cos 2C$

*$\cos 3A + \cos 3C + \cos 3C$

*$\cos(A\!-\!B)+\cos(B\!-\!C)+\cos(C\!-\!A)$

Now, I tried it simplifying using this formulae derived from Euler form of complex number: $\;2\cos x=e^{ix}+e^{-ix}\;$,
but nothing seems to work.
I do not understand, what is the significance of solving this question with complex numbers ?
 A: As noted by copper.hat, the conditions on cosine and sine imply that $$e^{iA} + e^{iB} + e^{iC} = 0.$$ These vectors all lie on the unit circle $|z| = 1$ in the complex plane. Thus, the sum of the first two vectors must be of length one, so as to be cancellable by the third. This can only occur if the angle between them is $2\pi/3$, or 120 degrees; see below for more detail. Thus $B = A \pm 2\pi/3$. Similarly, $C = A \pm 2\pi/3$, but $C$ cannot equal $B$ or the identity will not hold. Thus, we must have either $B = A + 2\pi/3$ and $C = A - 2\pi/3$, or the opposite. Since the expressions to be evaluated do not change when we swap $B$ and $C$, we may as well use the first choice.
To see in more detail why the angle between $A$ and $B$ must be $2\pi/3$, note that
\begin{align}
1 &= |-e^{iC}|^2 \\&= |e^{iA} + e^{iB}|^2 \\&= 2 + 2 \Re (e^{i(A - B)}),
\end{align}
from which it follows that $\cos(A - B) = -1/2.$ ($\Re$ means "real part.")
We can now answer the questions:

*

*\begin{align}
\cos{2A}+ \cos{2B} + \cos{2C} &= \Re (e^{2iA} + e^{2iB} + e^{2iC})\\ &= \Re(e^{2iA}(1 + e^{i4\pi/3} + e^{-i4\pi/3})) \\&= 0.
\end{align}


*\begin{align}\cos{3A}+ \cos{3B} + \cos{3C} &= \Re (e^{3iA} + e^{3iB} + e^{3iC}) \\&= 
\Re (e^{3iA} + e^{i(3A + 2\pi)} + e^{i(3A- 2\pi)}) \\&= 
3\cos{3A}.\end{align}


*\begin{align}\cos(A-B) + \cos(B-C) + \cos(C-A) &= \cos(-2\pi/3) + \cos(4\pi/3) + \cos(-2\pi/3) \\&= -3/2.\end{align}
